A function that packs consecutive duplicates of list elements into sublists in python

Question

What I mean is :

Eg., l = [1, 1, 1, 2, 2, 3, 3, 4]
pack(l) -> [[1,1,1], [2,2], [3,3], [4]]

I want to do this problem in a very basic way as I have just started i.e using loops and list methods.I have looked for other methods but they were difficult for me to understand

See here: https://stackoverflow.com/questions/5738901/removing-elements-that-have-consecutive-duplicates — Carsten, Apr 06 '20 at 14:23
What do you consider as "do this problem in a very basic way"? What have you tried so far? — MisterMiyagi, Apr 06 '20 at 14:23

score 2 · Answer 1 · answered Apr 06 '20 at 14:20

2

You can use groupby

from itertools import groupby

def pack(List):
    result = []
    for key,group in groupby(List):
         result.append(list(group))

    return result

l = [1, 1, 1, 2, 2, 3, 3, 4]
print(pack(l))

Or one-line:

l = [1, 1, 1, 2, 2, 3, 3, 4]
result = [list(group) for key,group in groupby(l)]
# [[1, 1, 1], [2, 2], [3, 3], [4]]

You can also use lambda to generate a function:

from itertools import groupby

l = [1, 1, 1, 2, 2, 3, 3, 4]

pack = lambda List: [list(group) for key,group in groupby(List)]
pack(l)

answered Apr 06 '20 at 14:20

jizhihaoSAMA

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I really appreciate this but I was talking about not using the above library.I want to use the basic approach as these things haven't been taught to us yet – Tanmay Apr 06 '20 at 14:21
@Tanmay So have you learned about `collections.Counter`? – jizhihaoSAMA Apr 06 '20 at 14:31

score 0 · Answer 2 · answered Apr 06 '20 at 14:18

0

This will work:

l = [1, 1, 1, 2, 2, 3, 3, 4]
l2 = [[entry] * l.count(entry) for entry in set(l)]
l2 # [[1, 1, 1], [2, 2], [3, 3], [4]]

answered Apr 06 '20 at 14:18

Carsten

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No. OP wants *consecutive duplicates*. Try with `[1, 1, 1, 2, 2, 3, 3, 4, 1]` – yatu Apr 06 '20 at 14:19

A function that packs consecutive duplicates of list elements into sublists in python

2 Answers2